(kx,ky,kz)=(0,0,0) solution for a free particle with PBC?

[Coffee_]

When dealing with Dirichlet boundary conditions, that is asking for the wavefunction to be exactly zero at the boundaries, it can be clearly seen that (0,0,0) is not a physical situation as it is not normalizable. (Wavefunction becomes just 0 then)

However when dealing with periodic boundary conditions, the basis is spanned by $e^{i\vec{k}.\vec{r}}$ where the only condition on $\vec{k}$ is that it has to correspond with $L^{3}$ cubic periodicity.

The problem now is that $\vec{k}=0$ does seem to give a non trivial solution with zero energy $\Psi=constant$ which is periodic and noralizable.

How do I interpret this werid 'constant' term in the general wavefunction part?

Sources:

Kittel eigth edition, p137

http://people.umass.edu/bvs/pbc.pdf

 

[Orodruin]

Why do you think it is weird?

 

[vanhees71]

To the contrary, it's not weird, but it's weird to think that with the Dirichlet boundary conditions you'd have a momentum operator. There is none, because there is no self-adjoint operator generating translations in this case. We have discussed this many times in this forum. Just search for it!

 

[Coffee_]

Oh I was thinking it was weird because I have never encountered such an extra constant term in the wavefunction so I was doubting my reasoning to arrive at the extra constant term. Based on your reactions I see that there is indeed nothing special about it, thanks.